702843. Balanced Array
Balanced Array
Slug: balanced-array07200720 Difficulty: Basic Id: 702843 Topic Tags: Arrays, Data Structures Company Tags: None
Summary
The Balanced Array problem is about finding a way to balance an array by moving elements from one end to the other. The goal is to distribute the elements evenly throughout the array, making sure that the difference between the maximum and minimum values is minimized.
Detailed Explanation
To solve this problem, we can start by sorting the array in ascending order. This will help us identify the range of values present in the array. Then, we can calculate the mean (average) value of the array. The idea is to move elements from one end of the array to the other until the difference between the maximum and minimum values is minimized.
Here's a step-by-step breakdown of the solution:
- Sort the array in ascending order.
- Calculate the mean (average) value of the array.
- Initialize two pointers,
leftandright, at the beginning and end of the array, respectively. - Compare the values at the
leftandrightpointers. If the value at theleftpointer is less than or equal to the mean, move theleftpointer one step forward. Otherwise, move therightpointer one step backward. - Repeat steps 3-4 until the difference between the maximum and minimum values of the array is minimized.
Here's an ASCII art diagram illustrating the process:
+---------------+
| mean |
+---------------+
|
v
+---------------+-----------+
| left | right|
| (small values) | (large values) |
+---------------+-----------+
Time complexity: O(n log n) due to sorting the array. Space complexity: O(1), as we only use a few extra variables.
Optimized Solutions
Java
public class BalancedArray {
public static void balanceArray(int[] arr) {
Arrays.sort(arr);
double mean = (double) Arrays.stream(arr).sum() / arr.length;
int left = 0, right = arr.length - 1;
while (Math.abs(mean - arr[left]) > Math.abs(mean - arr[right])) {
if (arr[left] <= mean) {
left++;
} else {
right--;
}
}
}
}